What is the center x^2+y^2+4x+4y=1 ?

The center x^2+y^2+4x+4y=1 is (-2,-2)

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center x^2+y^2+4x+4y=1

center

x^{2} + y^{2} + 4 x + 4 y = 1

(- 2, - 2)

Solution steps

x^{2} + y^{2} + 4 x + 4 y = 1

Rewrite

x^{2} + y^{2} + 4 x + 4 y = 1

in the form of the standard circle equation

(x - (- 2))^{2} + (y - (- 2))^{2} = 3^{2}

Therefore the circle properties are:

(a, b) = (- 2, - 2), r = 3

And the center is:

(- 2, - 2)

(x + 11)^{2} + (y + 8)^{2} = 147

(x - 5)^{2} + (y - 6)^{2} = 36

(x - 5)^{2} + (y + 7)^{2} = 9

x^{2} + y^{2} - 4 x - 6 y - 23 = 0

x^{2} + y^{2} - x + 2 y + 1 = 0